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Fix $n\geq 2$ and let $$\mathbb{H}^{n}=\mathbb{R}_{+}\times \mathbb{S}^{n-1}$$ be the hyperbolic space, so that any point $x\in \mathbb{H}^{n}$ can be represented in polar coordinates $x=(r, \theta)$, and equipped with the Riemannian metric $$g=dr^{2}+\sinh^{2}(r)d\theta^{2},$$ where $dr^{2}$ is the standard metric on $\mathbb{R}_{+}$ and $d\theta^{2}$ is the standard metric on the sphere $\mathbb{S}^{n-1}$. Denote with $\mu$ the Riemannian measure on $\mathbb{H}^{n}$ and with $\sigma$ the Riemannian measure of co-dimension $1$ on hypersurfaces on $\mathbb{H}^{n}$. It is a known fact that $\mathbb{H}^{n}$ admits the Isoperimetric inequality $$\sigma(\partial \Omega)\geq f(\mu(\Omega)),$$ for all precompact open sets $\Omega\subset \mathbb{H}^{n}$ with smooth boundary, where $f$ is defined by $$f(v)=c_{1}\left\{\begin{array}{cl}v, &v\geq 1\\ v^{\frac{n-1}{n}}, &v\leq 1,\end{array}\right.$$ for some small constant $c_{1}>0$.

Let us fix some precompact open set $U\subset \mathbb{H}^{n}$ and consider $\mathbb{H}^{n}\setminus U$ as a manifold with boundary $\partial U$.

Now my question:

Does the same Isoperimetric inequality now hold for precompact open sets $\Omega\subset \mathbb{H}^{n}\setminus U$ with smooth boundary (and possibly a smaller constant $c_{2}>0$) and if yes, where can I find a reference for this statement?

Specified question:

Let $U$ be a "nice" precompact open set, for example $U=B_{1}(o)$ is an open ball of radius $1$ for some point $o\in \mathbb{H}^{n}$. Does the same Isoperimetric inequality now hold for precompact open sets $\Omega\subset \mathbb{H}^{n}\setminus B_{1}(o)$ with smooth boundary (and possibly a smaller constant $c_{2}>0$)?

Note that we assume that, when considering a precompact open set $\Omega\subset \mathbb{H}^{n}\setminus U$ with smooth boundary, we have $\partial U\cap \partial \Omega=\emptyset$.

If this is not known for the hyperbolic space $\mathbb{H}^{n}$, is a similar statement known for other Riemannian manifolds, for example $\mathbb{R}^{n}$?

Thanks in advance for your help!

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  • $\begingroup$ What do you mean by zero Neumann boundary condition in this context? $\endgroup$ Jul 5, 2021 at 13:59
  • $\begingroup$ Assuming the answer to my first comment is that only the part of $\partial U$ on $\Omega$ is not counted, you cannot have a profile $f$ independent of $\Omega$ unless you impose some condition (bottlenecks could allow small perimeter and large volumes). Such an inequality does hold when $\Omega$ is convex. $\endgroup$ Jul 5, 2021 at 14:02
  • $\begingroup$ Yes, it is meant as you assume it in your answer. $\endgroup$
    – Shaq155
    Jul 5, 2021 at 14:33
  • $\begingroup$ Do you have a reference for that statement, when $\Omega$ is convex? $\endgroup$
    – Shaq155
    Jul 5, 2021 at 14:37
  • $\begingroup$ I specified my question. $\endgroup$
    – Shaq155
    Jul 5, 2021 at 14:52

1 Answer 1

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Some results are known under some restrictions. First, no uniform inequality can hold if $U$ is unrestrained: it could take the shape of (a neighborhood of) a bottle, and the bottleneck will make it possible to have $\Omega$ with large volume and small boundary. A natural restriction is to consider convex $U$ (so that the double of $\mathbb{H}^n$ still is negatively curved in the metric sense).

To sum up: assuming $U$ is convex, we even have sharp bounds. When $M=\mathbb{H}^n$ the dimensions $n=2,3,4$ are covered; when $M=\mathbb{R}^n$, you have a weaker inequality (the isoperimetric function has the form $f(v) = cv^{\frac{n-1}n}$, there is no linear asymptotic) but known in all dimensions.

Added in edit: I think that an equality $\sigma(\partial \Omega)\ge c\mu(\Omega)$ is true in all dimensions whenever $U$ is convex, and that the method in our paper with Kuperberg mentionned below can be used to prove this. The more precise bound for small $v$, I am not sure.

Note that $\mathbb{H}^n$ can even be replaced by a simply connected manifold $M$ of sectional curvature bounded above by some $\kappa\le 0$. In this setting:

  1. Choe proved in 2003 gave the sharp inequality when $M=\mathbb{R}^n$ and $U$ is a ball, as well as several other restricted cases: Relative isoperimetric inequality for domains outside a convex set. Archives Inequalities Appl 1 (2003): 241-250.

  2. In 2007, the general case of a convex set $U\subset \mathbb{R}^n$ was obtained by Choe, Ghomi and Ritoré. The relative isoperimetric inequality outside convex domains in $\mathbf{R}^n$. Calculus of Variations and Partial Differential Equations 29.4 (2007): 421-429.

  3. In 2006, Choe treated the case when $M$ has variable (nonpositive) curvature and dimension $4$. The double cover relative to a convex domain and the relative isoperimetric inequality. Journal of the Australian Mathematical Society 80.3 (2006): 375-382.

  4. In dimension $3$, the same was achieved by Choe and Ritoré in 2007. The relative isoperimetric inequality in Cartan-Hadamard 3-manifolds. J. für die reine und angewandte Mathematic (2007): 179-191. They also obtain the sharp inequality when $\kappa=-1$, in particular for $M=\mathbb{H}^3$.

  5. With Kuperberg, we gave a new proof of Choe's result from 2006, treating dimensions $2$ and $4$ and including (sharp) results when $M$ has curvature bounded above by $\kappa$ (either $0$ or negative). The Cartan–Hadamard conjecture and the Little Prince. Revista Matemática Iberoamericana 35.4 (2019): 1195-1258. Our results can be used to tackle some finite union of convex sets (you need to ensure that geodesic rays reflecting on the boundary of $U$ can only bounce a bounded number of time).

I may have missed other relevant references, but you should catch them by looking at papers citing the above ones.

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  • $\begingroup$ Thank you very much for your answer! Going through these papers I was wondering if in the case of $\mathbb{H}^{n}$, the profile $f$ is exactly as it I have written it in my Post with $f$ being linear for large volumes? $\endgroup$
    – Shaq155
    Jul 7, 2021 at 10:28
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    $\begingroup$ @Shaq155: yes, more precisely the profile (over all convex $U$) is $f(v)=\frac12 f_0(2v)$ where $f_0$ is the profile of $\mathbb{H}^n$, which is linear for large $v$. This comes from the optimal case, a half ball against a flat wall. $\endgroup$ Jul 8, 2021 at 9:34
  • $\begingroup$ @ Benoît Kloeckner: where exactly can I find the method which proves the linear asymptotic for large volumes in all dimensions in your paper with kuperberg or has there been made some progress by now? $\endgroup$
    – Shaq155
    Dec 4, 2021 at 11:20
  • $\begingroup$ @Shaq155: the relevant result in our paper with Kuperberg is Theorem 1.8; for this sharp bound, it is unfortunately necessary to digest most of the paper, which I admit is somewhat hard. But to prove a mere linear asymptotic, you can adapt the classical trick for the linear inequality in a Cartan-Hadamard manifold with curvature $\le -1$. The trick is to use Stokes equality on the gradient of either a Buseman function of the distance to a point. The adaptation for convex $U$ is to consider the double of $M\setminus U$ along $\partial U$, which is still CAT(-1). $\endgroup$ Dec 4, 2021 at 12:28

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